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If $\theta$ is an acute angle, simplify the expression: $$\frac{\sin\theta\cdot\cot\theta}{1-\sin^2\theta}$$

Solving: $$\begin{align} \frac{\sin\theta\cdot\cot\theta}{1-\sin^2\theta}&=\frac{\cot\theta \cdot\tan\theta}{\cos\theta} \\[4pt] &=\sec\theta \end{align}$$

My question:

Why did the problem specify that $\theta$ is acute? If the angle were not acute, then what is the solution?

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    $\begingroup$ If angle is obtuse, then $\cot \theta$ is negative because you are in the II quadrant. $\endgroup$
    – James
    May 2 '20 at 17:57
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    $\begingroup$ @James: $\sec\theta$ is also negative in II quadrant? $\endgroup$ May 2 '20 at 18:05
  • $\begingroup$ I mean it is sufficient to say simplify the trigonometric expression if you use trig identities , no conditions are changed for the final expression $\endgroup$ May 2 '20 at 18:12
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The simplified result $\sec \theta$ will hold for all values of $\theta$ where both the original expression and the final one are defined (and all intermediate steps in the manipulation remain valid with defined expressions)***.

The restriction to "acute angles" is presumably to avoid cases where those conditions fail. An acute angle is strictly less than $\frac {\pi}2$. If the angle were, say, $\pi$, then $\cot \theta$ (and the entire original expression) would not be defined.

Similarly, if $\theta = \frac{\pi}2$, you would run into a problem because the denominator would be $0$, and you can't divide by zero. The final expression $\sec \theta$ is also undefined for this value.

But acute angles also include $0$, and for this value $\cot \theta$ is undefined. So technically, they didn't restrict the domain sufficiently for rigour. They shouldn't have stopped with "acute angle", they should've said $\theta \in (0, \frac{\pi}2)$.

***With regard to the intermediate steps, you might be able to "hand wave" away undefined expressions here, but doing so rigorously involves bringing in the concept of limits, which is unreasonable in a simple trigonometric identity simplification.

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No condition of $\theta$ is required because you are using trigonometric identities which hold true for all the values of $\theta$

You will get the same solution as follows $$\frac{\sin\theta\cot\theta}{1-\sin^2\theta}$$ $$=\frac{\sin\theta\frac{\cos\theta}{\sin\theta}}{\cos^2\theta}$$ $$=\frac{1}{\cos\theta}$$ $$=\sec\theta$$

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  • $\begingroup$ But if $\theta=\frac{\pi}{2}$ it is undefined, isn't it? $\endgroup$ May 2 '20 at 18:07
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    $\begingroup$ obviously as the original expression is also undefined at that angle $\theta=\pi/2$. Therefore if original expression is undefined then final expression is also undefined at the same value of $\theta$. So no condition is necessary $\endgroup$ May 2 '20 at 18:08
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    $\begingroup$ The original expression is undefined when $$1-\sin^2\theta=0\iff \sin^2\theta=\sin^2\frac{\pi}{2}\iff \theta=n\pi\pm \frac{\pi}{2}$$ It is worth noticing that $\sec\theta$ is undefined at the same values i.e. $\theta=n\pi\pm \frac{\pi}{2}$ $\endgroup$ May 2 '20 at 18:16
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They probably said that $\theta$ is acute because any non right angle in a right angled triangle must be acute. I know the functions can be used for obtuse angles too but this is just my guess.

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